Forces on Particle, Find Third Force?

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving forces acting on a lemon on a frictionless table. The original poster describes the forces F1 and F2 with their magnitudes and angles, and seeks to determine a third force given different velocity conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the correct reference angles for the forces and question the calculations of the force components. There is discussion about the implications of static equilibrium and how to derive the third force from the given conditions.

Discussion Status

Participants are actively engaging with the problem, offering hints and clarifications about the relationships between forces and velocities. Some guidance has been provided regarding the interpretation of the forces and the implications of constant versus varying velocities.

Contextual Notes

There is mention of the mass of the lemon, but its relevance to the calculations is questioned. The problem involves multiple parts, each with different conditions regarding velocity, leading to varied interpretations of how to approach them.

GingerBread27
Messages
108
Reaction score
0
A lemon has three horizontal forces that act on it as it is on a frictionless table. Force F1 has a magnitude of 4.50 N and is at Angle1 = 26°. Force F2 has a magnitude of 7.00 N and is at Angle2= 26°. The lemon half has mass 0.0250 kg.
Picture is Attached.
What is the third force if the lemon half has the following velocities?


I did f1=4.5cos26i + 4.5sin26j
f2=7cos(-64)-7sin(-64)

THen I did f=(4.5cos26+7cos(-64))i+(-7sin(-64)+4.5sin36)j=7.11i+(-4.32j)

(a)At Zero velocity i got the I component to be -7.11 and the j component to be 4.32.
The j component is right but the i component is wrong and I don't know why.

(b)constant velocity v = (13.0 i - 14.0 j) m/s
(c)
varying velocity v = (13.0t i - 14.0t j) m/s, where t is time in seconds

What would I do to solve for B and C?
 

Attachments

  • W0083-N.jpg
    W0083-N.jpg
    2 KB · Views: 640
Last edited:
Physics news on Phys.org
Your reference angles may be wrong. I would make all angles with respect to the polar axis (positive x axis) so that 26 degree angle would become 116. I'm not sure if I understand what part a is asking. What is the force if the velocity is 0i+0j? Is that it?

Just a little hint for part c (part b is essentially part a over again): take the derivative of their varying velocity equation to get the constant vector acceleration
 
Last edited by a moderator:
yes that is what they are asking. I don't understand why you would use 116 as the reference angle, either way I still get the wronganswer. ANd I don't understand what you mean by part b being part a again.
 
Last edited:
Sorry. I meant angle 1 is (180-26) degrees with respect to the polar axis and angle 2 is (-90+26) degrees with respect to the polar axis. use those and you should get a correct answer.

Part b is exactly part A because from what I can see the velocity is still constant meaning no net force is being applied to it.
 
I don't see how using 154 as angle 1 would make the answers correct, since using angle1 as 26 degrees gave me the right j component.
 
Using it as an angle was correct because: sin(180-theta) = sin(theta) which you used in your j calculations correct? However, cos(180-theta) = -cos(theta) which messes you up on the i calculation. You basically found the force as if it was 26 degrees with respect to the polar axis. It is very important to choose the right angle.
 
so the i component would be (4.5cos154)+7cos(-64)?
 
I believe so.
 
It's wrong:( I don't understand. HOw can I get the J component and not the I!
 
  • #10
hang on I'll work out the question. Give me a moment. Edit: is the homework online? "so the i component would be (4.5cos154)+7cos(-64)?"

Ok I worked it out. the i component IS what you mentioned (approximately -0.975i) but remember this value is -F3*cos(theta) because F1*cos(theta)+F2*cos(theta)+F3*cos(theta) MUST equal 0.

Now work out the j component. That would be -F3*sin(theta). See if you can't figure out the two unknowns given the two equations.
 
Last edited by a moderator:
  • #11
online through webassign
 
  • #12
(see above edit)
 
  • #13
i don't understand
 
  • #14
What don't you understand specifically? You know since the object is stationary that it is in static equilibrium meaning that the sum of all the i components of the forces must equal zero as well as the sum of all the j components of the forces must equal zero (by Newton's second law). You found what the sum of forces F1 and F2 were but now you have to find F3 with components such that F1+F2+F3 = 0 in both the i and j directions. Manipulating the equation we find F1+F2 = -F3 or -(F1+F2) = F3.
 
Last edited by a moderator:
  • #15
so if that's the case then wouldn't the i component of the third vector be just .976?
 
  • #16
GingerBread27 said:
so if that's the case then wouldn't the i component of the third vector be just .976?

Yep it sure would.
 
  • #17
It was right, thank you so much. I'm sorry, but I just really don't understand this. Now for part b you said it was essentially the same as part a, so I thought you would just take f1+f2+f3=(13.0 i - 14.0 j) m/s but it's not working.
 
  • #18
GingerBread27 said:
It was right, thank you so much. I'm sorry, but I just really don't understand this. Now for part b you said it was essentially the same as part a, so I thought you would just take f1+f2+f3=(13.0 i - 14.0 j) m/s but it's not working.

Well consider what you're doing. 13.0i - 14.0j is the velocity right? You can't just set forces equal to velocity. What does a constant 13.0i-14.0j velocity tell you about the net force?
 
  • #19
the net force is constant?
 
  • #20
GingerBread27 said:
the net force is constant?

Well technically it is constant BUT consider what the velocity-time graph would look like. A straight line across would be it right? If acceleration is given by dv/dt, what is the acceleration?
 
  • #21
acceleration would be 0
 
  • #22
Ok so how's b different from a then? :) C shouldn't be too far of a leap from A so I think you can get it with the hint I gave you.
 
  • #23
B is the same as A, since F=ma and a is zero. lol. Great. I'm wondering here why the mass was given because i haven't used it to calculate anything yet. Now for C the lemon is moving with respect to time, so there must be acceleration, right? So the net force wouldn't be equal to zero. Then what am I not seeing? lol
 
  • #24
Right well if you have the velocity you know the acceleration since dv/dt = a
 
  • #25
I figured it out! :) THanks so much for your help :)
 

Similar threads

To find the resultant of forces on a lamina
  • · Replies 5 ·
Replies
5
Views
3K
How to Find the Path Function and Meeting Time for Three Moving Particles?
  • · Replies 14 ·
Replies
14
Views
2K
Trouble finding direction of two force vectors
  • · Replies 9 ·
Replies
9
Views
2K
Question about forces and acceleration
  • · Replies 3 ·
Replies
3
Views
3K
How Do Newton's Laws Apply to a Lemon Half on a Frictionless Surface?
  • · Replies 3 ·
Replies
3
Views
3K
Finding Other Force in 2-Force Problem
  • · Replies 5 ·
Replies
5
Views
2K
Conceptual question about Newton second law
  • · Replies 1 ·
Replies
1
Views
2K
How Do You Calculate the Third Force in a System of Connected Masses?
  • · Replies 6 ·
Replies
6
Views
2K
Calculating forces between stars?
  • · Replies 7 ·
Replies
7
Views
2K
Magnitude of resultant of three force as small as possible
  • · Replies 1 ·
Replies
1
Views
4K